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a^3 +c^3 = (a+c). (a^2 -a.c+c^2)
= (a+c)^3 -3 ac.(a+c)
=> a^3+c^3-3abc+b^3 =(a+c)^3-3ac (a+c)-3abc +b^3
=(a+c)^3+b^3 -3ac (b+(a+c))
=(a+c+b). ((a+c)^2-(a+c).b+b^2) -3ac (a+c+b)
=(a+c+b)^3-3(a+c)b. (a+c+b)-3ac (a+c+b)
=(a+c+b)((a+c+b)^2 -3ab-3bc-3ac) (1)
(a-b)^2 + (b-c)^2 +(a-c)^2
= 2a^2 +2b^2+2c^2 -2ab-2bc-2ac
=2 (a^2+b^2+c^2-ac-ab-bc)
=2((a+b)^2-3ab +c^2 -ac-bc)
=2 ((a+b+c)^2-2(ac+bc)-3ab-ac-bc)
=2 (( a+c+b)^2 -3ab-3bc -3ac) (2)
Từ (1),(2) =>(a^3+b^3+c^3-3abc)/((a-b)^2
+(b-c)^2+(c-a)^2)
=(a+b+c)/2
a) \(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)
\(=\frac{a^2b-a^2c+b^2c-b^2a+c^2\left(a-b\right)}{ab^2-b^3-ac^2+bc^2}\)
\(=\frac{\left(a^2b-b^2a\right)+\left(b^2c-a^2c\right)+c^2\left(a-b\right)}{b^2\left(a-b\right)-c^2\left(a-b\right)}\)
\(=\frac{ab\left(a-b\right)+c\left(b^2-a^2\right)+c^2\left(a-b\right)}{\left(b^2-c^2\right)\left(a-b\right)}\)
\(=\frac{ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)}{\left(b-c\right)\left(b+c\right)\left(a-b\right)}\)
\(=\frac{ab-c\left(a+b\right)+c^2}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{ab-ac+c^2-bc}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{a\left(b-c\right)-c\left(b-c\right)}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{\left(b-c\right)\left(a-c\right)}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{a-b}{b+c}\)
Sửa đề: \(P=\frac{a^3+b^3+c^3-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(P=\frac{a^3+b^3+c^3-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(P=\frac{\left(a+b\right)^3+c^3-3abc-3a^2b-3ab^2}{a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2}\)
\(P=\frac{\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right).c+c^2\right]-3ab\left(a+b+c\right)}{2.\left(a^2+b^2+c^2-ab-bc-ca\right)}\)
\(P=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc+3ab\right)}{2.\left(a^2+b^2+c^2-ab-bc-ca\right)}\)
\(P=\frac{5\left(a^2+b^2+c^2-ab-ac-bc\right)}{2.\left(a^2+b^2+c^2-ab-bc-ca\right)}\)( a+b+c=0)
\(P=\frac{5}{2}\left[\left(a^2+b^2+c^2-ab-bc-ca\right)\ne0\right]\)
Ta có \(P=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{a^2-ab+b^2+b^2-bc+c^2+c^2-ac+a^2}\)
\(=\frac{5\left(...\right)}{2\left(...\right)}=\frac{5}{2}\)
phân tích tử thức:
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Phân tích mẫu thức:\(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(ab^2-a^2b+bc^2-b^2c+ca^2-c^2a\right)\)
\(=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(\Rightarrow A=\frac{3\left(a^2+b^2+c^2-ab-bc-ca\right)}{3\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{a^2+b^2+c^2-ab-bc-ca}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
a 3 + b 3 + c 3 = 3abc⇔a 3 + b 3 + c 3 − 3abc = 0
⇔ a + b 3 − 3ab a + b + c 3 − 3abc = 0
⇔ a + b 3 + c 3 − 3ab a + b + 3abc = 0
⇔ a + b + c a 2 + b 2 + c 2 + 2ab − ac − bc − 3ab a + b + c = 0
⇔ a + b + c a 2 + b 2 + c 2 − ab − bc − ac = 0
⇔ 2 a + b + c a − b 2 + b − c 2 + c − a /2 = 0
Vì a,b,c > 0 nên a+b+c > 0
Do đó : a − b 2 = 0
b − c 2 = 0
c − a 2 = 0
⇒a = b = c
k cho mk nha
Sửa đề cho nó đẹp
\(\frac{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}\)
\(=\frac{3\left(a-b\right)\left(a-c\right)\left(c-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=-3\)
Phân tích mẫu thức thành nhân tử :
\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)+b^2c-ab^2+ac^2-bc^2\)
\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b^2-c^2\right)\)
\(=\left(b-c\right)\left(a^2+bc-ab-ac\right)\)
\(=\left(b-c\right)\left[a\left(a-b\right)-c\left(a-b\right)\right]=\left(b-c\right)\left(a-c\right)\left(a-b\right).\)
Do đó : \(A=\frac{\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b\right)^3}{-\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Nhận xét : Nếu \(x+y+z=0\) thì \(x^3+y^3+z^3=3xyz.\)
Đặt \(b-c=x,c-a=y,a-b=z\) thì \(x+y+z=0\)
Theo nhận xét trên : \(A=\frac{x^3+y^3+z^3}{-xyz}=\frac{3xyz}{-xyz}=-3.\)
Tử:
(b - c)3 + (c - a)3 + (a - b)3
= (b - c + c - a + a - b)3 - 3(b - c + c - a)(b - c + a - b)(c - a + a - b)
= 0 - 3(b - a)(a - c)(c - b)
= 3(a - b)(a - c)(c - b)
Mẫu:
a2(b - c) + b2(c - a) + c2(a - b)
= a2(b - c) + b2c - ab2 + ac2 - bc2
= a2(b - c) - a(b2 - c2) + bc(b - c)
= a2(b - c) - a(b - c)(b + c) + bc(b - c)
= (b - c)(a2 - ab - ac + bc)
= (b - c)[a(a - b) - c(a - b)]
= (b - c)(a - b)(a - c)
\(A=\frac{3\left(a-b\right)\left(a-c\right)\left(c-b\right)}{\left(b-c\right)\left(a-b\right)\left(a-c\right)}\)
\(=\frac{3\left(c-b\right)}{b-c}\)
a: \(=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)-3abc}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\dfrac{\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)}{a^2+b^2+c^2-ab-bc-ac}\)
=a+b+c
b:
Sửa đề: \(=\dfrac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y\right)^3+z^3+3xy\left(x-y\right)+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y+z\right)\left(x^2-2xy+y^2-xz+yz+z^2\right)+3xy\left(x-y+z\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)
\(=\dfrac{\left(x-y+z\right)\left(x^2+y^2+z^2+xy-xz+yz\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)
\(=\dfrac{x-y+z}{2}\)
a) \(\dfrac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ca}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{a^2+b^2+c^2-ab-bc-ca}\)
\(=a+b+c\)
\(\frac{a^3+b^3-c^3+3abc}{\left(a-b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2}=\frac{\left(a+b\right)^3-c^3-3ab\left(a+b\right)+3abc}{2a^2+2b^2+2c^2-2ab+2bc+2ac}\)
\(=\frac{\left(a+b-c\right)\left[\left(a+b\right)^2+c\left(a+b\right)+c^2\right]-3ab\left(a+b-c\right)}{2a^2+2b^2+2c^2-2ab+2bc+2ac}\)
\(=\frac{\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2-3ab\right)}{2\left(a^2+b^2+c^2-ab+bc+ac\right)}\)
\(=\frac{a+b-c}{2}\)